Question

The resistivity of the silver is $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{-8}\mathbf{\Omega }\mathbf{\cdot }\mathbf{m}$at room temperature of (300 K), while that of silicon is about$\mathbf{10}\mathbf{\Omega }\mathbf{\cdot }\mathbf{m}$

(a) Show that this disparity follows, at least to a rough order of magnitude from the approximate 1 eV band gap in silicon.

(b) What would you expect for the room temperature resistivity of diamond, which has a band gap of about 5 eV.

Short answer

(a)The factor is approximately similar to the ratio between the resistivity.

(b) The resistivity for the diamond is$1.6\times {10}^{34}\Omega \cdot \text{m}$ .

Step-by-step solution

Determine the formulas

Consider the exponential formula as per the Boltzmann’s distribution:

${\mathbf{e}}^{\mathbf{-}\frac{{\mathbf{E}}_{\mathrm{gap}}}{\mathbf{2}{\mathbf{k}}_b\mathbf{T}}}$ ….. (1)

Here,$k_b$ is Boltzmann constant and the temperature is T.

Consider the expression for the vibrational distance as:

Determine the answer for part (a).

Substitute the values in the equation () and solve as:

$\begin{array}{c}{e}^{\frac{-E_{gap}}{2k_{b}T}}=e^{\frac{-1\text{\hspace{0.17em}eV}}{2\times 1.38\times {10}^{-23}\times 300}}\text{\hspace{0.17em}eV}\\ =\text{4}\times {10}^{-9}\text{\hspace{0.17em}eV}\end{array}$

Note the availability of the electron is less than $\text{4}\times {10}^{-9}\text{\hspace{0.17em}eV}$so the resistivity is approximately$\frac{1}{\text{4}\times {10}^{-9}\text{\hspace{0.17em}eV}}=2.5\times {10}^8$

Consider the equation for the ration between the resistivity of the silicon and the silver as:

$\frac{{\rho }_{silicon}}{{\rho }_{silver}}$

Substitute the values and solve as:

$\begin{array}{c}\frac{{\rho }_{silicon}}{{\rho }_{silver}}=\frac{1.6\times {10}^{-8}\Omega \cdot \text{m}}{10\Omega \cdot \text{m}}\\ =1.6\times {10}^{-9}\end{array}$

Therefore, the factor is approximately similar to the ratio between the resistivity.

Determine the answer for part (b).

Since, the energy gap of the diamond is$-5\text{\hspace{0.17em}eV}$ .

Solve for the gap as:

$\begin{array}{c}{E}_{gap}=-5\text{\hspace{0.17em}eV}\\ =(-5\text{\hspace{0.17em}eV})\left(\frac{1.6\times {10}^{-19}\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}eV}}\right)\\ =(-5)(1.6\times {10}^{-19}\text{\hspace{0.17em}J})\end{array}$

Consider the formula for the resistivity of the diamond as:

${\rho }_{diamond}=e^{\frac{-E_{gap2}}{2k_{b}T}}\frac{{\rho }_{silicon}}{{\rho }_{silver}}$

Resolve the equation as:

${\rho }_{diamond}=\frac{{\rho }_{silver}}{e^{\frac{-E_{gap2}}{2k_{b}T}}}$

Substitute the values and solve as:

$\begin{array}{c}{\rho }_{diamond}=\frac{1.6\times {10}^{-8}\Omega \cdot \text{m}}{e^{\frac{-\left(5\right)(1.6\times {10}^{-19}\text{\hspace{0.17em}J})}{2\times 1.38\times {10}^{-23}\times 300}}}\\ =\frac{1.6\times {10}^{-8}\Omega \cdot \text{m}}{{10}^{-42}}\\ =1.6\times {10}^{-8}\Omega \cdot \text{m}\left({10}^{42}\right)\\ =1.6\times {10}^{34}\Omega \cdot \text{m}\end{array}$

Therefore, the resistivity of the diamond is$1.6\times {10}^{34}\Omega \cdot \text{m}$ .